标签：
https://leetcode.com/problems/linked-list-cycle-ii/
Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
Follow up:
Can you solve it without using extra space?
题目：如果链表存在环路，找到环路的起始结点。
注：参加博客http://www.cnblogs.com/zuoyuan/p/3701877.html 写的很详细
经过推到得知，当slow从head出发，fast从相遇点出发，一步一步前进，它们会再次相遇在起始结点。
1 # Definition for singly-linked list.
2 # class ListNode:
3 # def __init__(self, x):
4 # self.val = x
5 # self.next = None
6
7 class Solution:
8 # @param head, a ListNode
9 # @return a list node
10 def detectCycle(self, head):
11 if head==None or head.next==None:
12 return None
13 slow=fast=head
14 while fast and fast.next: #找到slow和fast的相遇点，让fast待在那里
15 slow=slow.next
16 fast=fast.next.next
17 if slow==fast:
18 break
19 if slow==fast: #不知前面跳出循环的原因是不满足while条件(无环),还是满足if而break(找到相遇点),所以要判断
20 slow=head #让slow回到head
21 while slow!=fast:
22 slow=slow.next #fast和slow同时走直到while不满足（即相遇）
23 fast=fast.next
24 return slow #再次相遇点即为起点，返回该点
25 return None #对应if slow==fast的else情况，即无环，返回None
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原文地址：http://www.cnblogs.com/lzsjy421/p/4612702.html
